Problem: $\text E = \left[\begin{array}{rr}-1 & 3 \\ 0 & 4\end{array}\right]$ and $\text B = \left[\begin{array}{rrr}0 & 5 & -2 \\ 5 & 2 & 2\end{array}\right]$ Let $\text {H = EB}$. Find $\text H$. $ {H = }$
Solution: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{B}$. $ \text {H}=\left[\begin{array}{rr}{-1} & {3} \\ 0 & 4\end{array}\right]\left[\begin{array}{rr} {0} & 5 & -2 \\ {5} & 2 & 2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(-1,3)\cdot(0,5)\\\\ &=-1 \cdot 0 + 3\cdot 5\\\\ &=15 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0 \cdot 0 + 4\cdot 5 = 20$ (Choice B) B $-1 \cdot 5 + 3\cdot 2 = 1$ (Choice C) C $0 \cdot 5 + 4\cdot 2 = 8$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}= \left[\begin{array}{rrr}15 & 1 & 8 \\ 20 & 8 & 8\end{array}\right]$